Find the gradient of $f(x, y, z)=\left(x^{2}+y^{2}+z^{2}\right)^{-1 / 2}+\ln (x y z)$ at the point $(2,1,2)$. \[ \left.\nabla f\right|_{(2,1,2)}= \] \[ k \] (Type integers or simplified fractions.)

Step 1 ：Given the function \(f(x, y, z)=(x^{2}+y^{2}+z^{2})^{-1 / 2}+\ln (x y z)\), we need to find the gradient at the point (2,1,2).

Step 2 ：The gradient of a function is a vector of its partial derivatives with respect to each variable. In this case, we need to find the partial derivatives of the function with respect to x, y, and z, and then evaluate these at the point (2,1,2).

Step 3 ：The partial derivative of the function with respect to x is given by the derivative of the function with respect to x, holding y and z constant. Similarly, the partial derivatives with respect to y and z are found by differentiating with respect to y and z, holding the other variables constant.

Step 4 ：The partial derivative of the function with respect to x is: \(\frac{\partial f}{\partial x} = -\frac{x}{\left(x^{2}+y^{2}+z^{2}\right)^{3 / 2}} + \frac{1}{x}\)

Step 5 ：Similarly, the partial derivatives with respect to y and z are: \(\frac{\partial f}{\partial y} = -\frac{y}{\left(x^{2}+y^{2}+z^{2}\right)^{3 / 2}} + \frac{1}{y}\) and \(\frac{\partial f}{\partial z} = -\frac{z}{\left(x^{2}+y^{2}+z^{2}\right)^{3 / 2}} + \frac{1}{z}\)

Step 6 ：We can then substitute the point (2,1,2) into these expressions to find the gradient at this point.

Step 7 ：The partial derivative of the function with respect to x at the point (2,1,2) is: 0.425925925925926

Step 8 ：The partial derivative of the function with respect to y at the point (2,1,2) is: 0.962962962962963

Step 9 ：The partial derivative of the function with respect to z at the point (2,1,2) is: 0.425925925925926

Step 10 ：Final Answer: The gradient of \(f(x, y, z)=(x^{2}+y^{2}+z^{2})^{-1 / 2}+\ln (x y z)\) at the point (2,1,2) is \(\boxed{(0.425925925925926, 0.962962962962963, 0.425925925925926)}\)